I thought I should expand on my comments under @gung's answer, explain the relevance of the ratio I mentioned in comments and show some of the interesting values.
The Welch-Satterthwaite d.f. are
$$\nu_{_W} = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{s_1^4}{n_1^2\nu_1}+\frac{s_2^4}{n_2^2\nu_2}}$$
Note that $s_i^2/n_i$ is the estimated variance of the $i^\text{th}$ sample mean. Let $r_i=s_i^2/n_i$, $i=1,2$ and $r=r_1/r_2$, so
\begin{align} \nu_{_W} &= \frac{\left(r_1+r_2\right)^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \newline \newline &=\frac{\left(r_1+r_2\right)^2}{r_1^2+r_2^2}\frac{r_1^2+r_2^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \newline \newline &=\frac{\left(r+1\right)^2}{r^2+1}\frac{r_1^2+r_2^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \end{align}
The first term is $1+\text{sech}(\log(r))$, which increases from $1$ at $r=0$ to $2$ at $r=1$ and then decreases to $1$ at $r=\infty$; it's symmetric in $\log r$.
The second term is a weighted harmonic mean:
$$H(\underline{x})=\frac{\sum_{i=1}^n w_i }{ \sum_{i=1}^n \frac{w_i}{x_i}}\,.$$
of the d.f., where $w_i=r_i^2$ are the relative weights to the two d.f.
Which is to say, when $r_1/r_2$ is very large, it converges to $\nu_1$. When $r_1/r_2$ is very close to $0$ it converges to $\nu_2$. When $r_1=r_2$ you get twice the harmonic mean of the d.f., and when $s_1^2=s_2^2$ you get the usual equal-variance t-test d.f., which is also the maximum possible value for $\nu_{_W}$.
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With an equal-variance t-test, if the assumptions hold, the square of the denominator is a constant times a chi-square random variate.
The square of the denominator of the Welch t-test isn't (a constant times) a chi-square; however, it's often not too bad an approximation. A relevant discussion can be found here